[Mondial]

There is no voltage at the emitter of Q17 because the transistor is not conducting and there is no current through R97. For Q17 to turn on there must be at least .6 V between base and emitter. Since you have only .5 V at the base, the transistor is not turned on and no current flows. Operating properly, with 1.0 V at the base, there will be .6 V across the BE junction, and the resulting current flow through R97 will produce the .4 V voltage drop across it.

What is the corresponding voltage across R95, Q15's emitter resistor? Also measure the voltage across R89, as there should also be about 1.0 V across this resistor to turn on Q15 and produce current flow.

The current through Q15 and Q17 should be equal, which will provide identical voltage drops across R97 and R95 .